Question

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60 $\times$ 10${}^7$ V>m. The capacitor is to have a capacitance of 1.25 $\times$ 10${}^{-9}$ F and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

Short answer

The minimum plate area needed is approximately 0.0136 m².

Step-by-step solution

Understand the Problem

We need to find the minimum area of the plates of a parallel-plate capacitor given specific electrical properties: dielectric constant, dielectric strength, capacitance, and maximum potential difference.

Use the Capacitance Formula

Recall the formula for the capacitance of a parallel-plate capacitor: $$C=\frac{{\epsilon }_0{\epsilon }_{r}A}{d}$$where ${\epsilon }_0$ is the permittivity of free space ($8.85\times {10}^{-12}\text{ F/m}$), ${\epsilon }_r$ is the relative permittivity (3.60), $A$ is the area, and $d$ is the distance between the plates. Here, $C=1.25\times {10}^{-9}\text{ F}$.

Apply Dielectric Strength Condition

The dielectric strength provides the maximum electric field ($E_{max}$) the material can withstand, given by $E_{max}=1.60\times {10}^7\text{ V/m}$. The potential difference $V$ is related to the electric field by $V=E_{max}\cdot d$. Solve for $d$:$$d=\frac{V}{E_{max}}=\frac{5500\text{V}}{1.60\times {10}^7\text{V/m}}\approx 3.44\times {10}^{-4}\text{m}.$$

Rearrange the Capacitance Formula for Area

We know $C$, ${\epsilon }_0$, ${\epsilon }_r$, and $d$, and we can solve for $A$:$$A=\frac{C\cdot d}{{\epsilon }_0{\epsilon }_r}=\frac{1.25\times {10}^{-9}\text{F}\cdot 3.44\times {10}^{-4}\text{m}}{8.85\times {10}^{-12}\text{F/m}\cdot 3.60}\approx 1.36\times {10}^{-2}{\text{m}}^2.$$

Solution Conclusion

The calculation yields the minimum area of the capacitor plates necessary to meet the given requirements.

Key concepts

Dielectric Strength

Dielectric strength is a critical property of insulating materials, especially ones used in capacitors. It refers to the maximum electric field that a material can withstand without breaking down or losing its insulating properties. In simple terms, it's the material's ability to resist an electric breakdown.
For a dielectric material, the dielectric strength is measured in volts per meter (V/m). A higher dielectric strength means that the material can endure a stronger electric field. This is essential in preventing breakdown in capacitors, ensuring the device functions correctly under high voltages.
In the context of capacitor design, knowing the dielectric strength helps determine how thin the material can be without risking failure. Choosing a material with a suitable dielectric strength ensures that the capacitor can operate safely at the intended voltage, as calculated in our exercise.

Parallel-Plate Capacitor

A parallel-plate capacitor is a straightforward type of capacitor that features two conducting plates separated by an insulating material called a dielectric. This setup is widely used due to its simplicity and effectiveness.
The capacitance of this kind of capacitor is determined by the area of the plates, the separation distance between the plates, and the properties of the dielectric in between. Larger plate areas and smaller separation distances increase capacitance, making for a more efficient energy storage device.
Parallel-plate capacitors are commonly used in electronic circuits and can be found in devices ranging from radios to laptops. Understanding the factors that influence their design, like the dielectric material and plate area, is key to optimizing their performance for various applications.

Capacitance Formula

The capacitance formula for a parallel-plate capacitor is: $$C=\frac{{\epsilon }_0{\epsilon }_{r}A}{d}$$where:

• $C$ is the capacitance in farads (F).
• ${\epsilon }_0$ is the permittivity of free space ($8.85\times {10}^{-12}$ F/m).
• ${\epsilon }_r$ is the relative permittivity or dielectric constant of the material.
• $A$ is the area of the plates.
• $d$ is the separation distance between the plates.

This formula calculates how much electric charge a capacitor can store. The greater the capacitance, the more charge it can hold for a given voltage.
By rearranging the formula, we can find different variables, such as the minimum area of the plates required for a specific capacitance, as done in the exercise. This is crucial for designing capacitors that meet precise electrical criteria.